Further Maths Momentum and impulse Practice Questions

Free Further Maths Momentum and impulse practice questions with full step-by-step worked solutions. Covers impulse-momentum-principle, impulse, sign-convention, impulse-of-a-constant-force. Practise exam-style problems and check your method.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A particle is moving in a straight line on a smooth horizontal plane and is modelled as a particle. Take the direction of motion of the particle immediately before the impulse acts as the positive direction. The mass of the particle is m=0.5 kgm=0.5\ \text{kg}. Its velocity immediately before it receives an impulse is u=8 m s−1u=8\ \text{m s}^{-1} and its velocity immediately afterwards is v=−2 m s−1v=-2\ \text{m s}^{-1}. Find II, the impulse received by the particle, in N s\text{N s}.
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Worked solution

  1. State the impulse-momentum principle

    I=mv−mu=m(v−u)I=mv-mu=m\left(v-u\right)

    The impulse acting on the particle equals its change in momentum.

  2. Substitute the given values

    I=0.5×(−2)−0.5×8I=0.5\times\left(-2\right)-0.5\times8

    The mass is multiplied by each velocity, keeping the signs.

  3. State the final answer with its units

    I=−5 N sI=-5\ \text{N s}

    This is the impulse, quoted with its units.

Answer
I=−5 N sI=-5\ \text{N s}
Question 2
2 markseasy
A particle of mass mm is moving in a straight line and is modelled as a particle. Take the direction of motion of the particle immediately before the impulse acts as the positive direction. The particle receives an impulse I\mathbf{I} which changes its velocity from u\mathbf{u} to v\mathbf{v}. Which of the following expressions gives I\mathbf{I}?
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Worked solution

  1. State the impulse-momentum principle

    I=mv−mu\mathbf{I}=m\mathbf{v}-m\mathbf{u}

    The impulse on a particle equals its change in momentum.

  2. Note the order of the two terms

    I=m(v−u)\mathbf{I}=m\left(\mathbf{v}-\mathbf{u}\right)

    It is the final momentum minus the initial momentum, not the other way round.

  3. Select the correct expression

    I=mv−mu\mathbf{I}=m\mathbf{v}-m\mathbf{u}

    This is the only option that follows from the principle being used.

Answer
I=mv−mu\mathbf{I}=m\mathbf{v}-m\mathbf{u}
Question 3
4 marksintermediate
Two small spheres AA and BB, which are modelled as particles, are moving in the same straight line on a smooth horizontal plane. Take the direction of motion of AA immediately before the collision as the positive direction. The masses of AA and BB are mAm_{A} and mBm_{B}, and their velocities immediately before the collision are uAu_{A} and uBu_{B}. The spheres collide and coalesce, and immediately afterwards the combined particle has velocity vv. Which of the following expressions gives vv?
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Worked solution

  1. State the principle of conservation of linear momentum

    mAuA+mBuB=mAvA+mBvBm_{A}u_{A}+m_{B}u_{B}=m_{A}v_{A}+m_{B}v_{B}

    The plane is smooth and the impact is instantaneous, so no external impulse acts on the system and the total momentum is unchanged.

  2. Use the fact that the spheres coalesce

    vA=vB=vv_{A}=v_{B}=v

    After coalescing the two spheres move as one particle.

  3. Write the total momentum after the collision

    pafter=(mA+mB)vp_{\text{after}}=\left(m_{A}+m_{B}\right)v

    The combined particle has mass mA+mBm_{A}+m_{B}.

  4. Equate the momenta

    mAuA+mBuB=(mA+mB)vm_{A}u_{A}+m_{B}u_{B}=\left(m_{A}+m_{B}\right)v

    No external impulse acts, so the total momentum is conserved.

  5. Reject the option with a difference of momenta

    mAuA−mBuB≠pbeforem_{A}u_{A}-m_{B}u_{B}\neq p_{\text{before}}

    The signs of the velocities already record the directions.

  6. Select the correct expression

    v=mAuA+mBuBmA+mBv=\frac{m_{A}u_{A}+m_{B}u_{B}}{m_{A}+m_{B}}

    This is the only option that follows from the principle being used.

Answer
v=mAuA+mBuBmA+mBv=\frac{m_{A}u_{A}+m_{B}u_{B}}{m_{A}+m_{B}}
Question 4
6 markshard
Two small spheres AA and BB, which are modelled as particles, are moving in the same straight line on a smooth horizontal plane. Take the direction of motion of AA immediately before the collision as the positive direction. The masses of AA and BB are mAm_{A} and mBm_{B}, their velocities immediately before the collision are uAu_{A} and uBu_{B}, and their velocities immediately afterwards are vAv_{A} and vBv_{B}. Which of the following expressions gives vBv_{B}?
Show worked solution

Worked solution

  1. State the principle of conservation of linear momentum

    mAuA+mBuB=mAvA+mBvBm_{A}u_{A}+m_{B}u_{B}=m_{A}v_{A}+m_{B}v_{B}

    The plane is smooth and the impact is instantaneous, so no external impulse acts on the system and the total momentum is unchanged.

  2. State the positive direction being used

    positive→direction of motion of A before impact\text{positive}\rightarrow\text{direction of motion of }A\text{ before impact}

    Velocity and impulse are vectors, so a direction must be fixed before any numbers are written down.

  3. Isolate the term containing vBv_{B}

    mBvB=mAuA+mBuB−mAvAm_{B}v_{B}=m_{A}u_{A}+m_{B}u_{B}-m_{A}v_{A}

    Subtract the momentum of AA after the collision from the total.

  4. Divide by mBm_{B}

    vB=mAuA+mBuB−mAvAmBv_{B}=\frac{m_{A}u_{A}+m_{B}u_{B}-m_{A}v_{A}}{m_{B}}

    This gives the velocity of BB after the collision.

  5. Reject the option that forgets to divide by the mass

    mAuA+mBuB−mAvA=mBvB≠vBm_{A}u_{A}+m_{B}u_{B}-m_{A}v_{A}=m_{B}v_{B}\neq v_{B}

    That expression is a momentum, not a velocity.

  6. Reject the option that drops the initial momentum of BB

    mA(uA−vA)mB=vB−uB\frac{m_{A}\left(u_{A}-v_{A}\right)}{m_{B}}=v_{B}-u_{B}

    That expression is the change in the velocity of BB, not its final velocity.

  7. State the impulse-momentum principle in vector form

    I=mv−mu\mathbf{I}=m\mathbf{v}-m\mathbf{u}

    The impulse acting on a body is equal to the change in its momentum.

  8. Recall that momentum is a vector

    p=mv\mathbf{p}=m\mathbf{v}

    Momentum points in the same direction as the velocity, so signs matter.

  9. Note the units of impulse and momentum

    1 N s=1 kg m s−11\ \text{N s}=1\ \text{kg m s}^{-1}

    Impulse and momentum have the same dimensions; either unit may be quoted.

  10. State Newton's third law for impulses

    IA=−IB\mathbf{I}_{A}=-\mathbf{I}_{B}

    The impulses the two bodies exert on each other are equal and opposite.

  11. Select the correct expression

    vB=mAuA+mBuB−mAvAmBv_{B}=\frac{m_{A}u_{A}+m_{B}u_{B}-m_{A}v_{A}}{m_{B}}

    This is the only option that follows from the principle being used.

Answer
vB=mAuA+mBuB−mAvAmBv_{B}=\frac{m_{A}u_{A}+m_{B}u_{B}-m_{A}v_{A}}{m_{B}}
Question 5
9 markschallenging
Two small spheres AA and BB, which are modelled as particles, are moving in the same straight line on a smooth horizontal plane. Take the direction of motion of AA immediately before the collision as the positive direction. The masses of AA and BB are mAm_{A} and mBm_{B}, their velocities immediately before the collision are uAu_{A} and uBu_{B}, and their velocities immediately afterwards are vAv_{A} and vBv_{B}. Which of the following expressions gives IBI_{B}, the impulse exerted on BB by AA during the collision?
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Worked solution

  1. State the impulse-momentum principle for BB

    IB=mBvB−mBuBI_{B}=m_{B}v_{B}-m_{B}u_{B}

    The impulse on BB equals the change in the momentum of BB.

  2. Factorise out the mass

    IB=mB(vB−uB)I_{B}=m_{B}\left(v_{B}-u_{B}\right)

    The mass is a common factor of the two momenta.

  3. State the positive direction being used

    positive→direction of motion of A before impact\text{positive}\rightarrow\text{direction of motion of }A\text{ before impact}

    Velocity and impulse are vectors, so a direction must be fixed before any numbers are written down.

  4. Note the equivalent expression from Newton's third law

    IB=−IA=−mA(vA−uA)I_{B}=-I_{A}=-m_{A}\left(v_{A}-u_{A}\right)

    Conservation of momentum makes the two expressions equal in value.

  5. Reject the option with the terms reversed

    mB(uB−vB)=−IBm_{B}\left(u_{B}-v_{B}\right)=-I_{B}

    Reversing the order of the two momenta reverses the direction of the impulse.

  6. Reject the option without a mass

    vB−uB≠IBv_{B}-u_{B}\neq I_{B}

    A change of velocity is not an impulse.

  7. State the impulse-momentum principle in vector form

    I=mv−mu\mathbf{I}=m\mathbf{v}-m\mathbf{u}

    The impulse acting on a body is equal to the change in its momentum.

  8. Recall that momentum is a vector

    p=mv\mathbf{p}=m\mathbf{v}

    Momentum points in the same direction as the velocity, so signs matter.

  9. Note the units of impulse and momentum

    1 N s=1 kg m s−11\ \text{N s}=1\ \text{kg m s}^{-1}

    Impulse and momentum have the same dimensions; either unit may be quoted.

  10. State Newton's third law for impulses

    IA=−IB\mathbf{I}_{A}=-\mathbf{I}_{B}

    The impulses the two bodies exert on each other are equal and opposite.

  11. Note why momentum is conserved here

    ∑Iexternal=0\sum\mathbf{I}_{\text{external}}=\mathbf{0}

    During the instantaneous interaction the only impulses are internal to the system.

  12. State the impulse of a constant force

    I=Ft\mathbf{I}=\mathbf{F}t

    A constant force acting for a time tt delivers an impulse Ft\mathbf{F}t.

  13. Record the modelling assumptions

    particles; smooth plane; instantaneous impact\text{particles; smooth plane; instantaneous impact}

    Modelling the bodies as particles removes rotation, and a smooth plane gives no frictional impulse.

  14. Check the sign convention

    positive=the direction stated in the question\text{positive}=\text{the direction stated in the question}

    Every velocity and every impulse must be measured with the same positive direction.

  15. Note that a negative answer is a physical result

    v<0  ⟺  motion in the negative directionv<0\iff\text{motion in the negative direction}

    A negative velocity simply means the body moves opposite to the chosen positive direction.

  16. Select the correct expression

    IB=mB(vB−uB)I_{B}=m_{B}\left(v_{B}-u_{B}\right)

    This is the only option that follows from the principle being used.

Answer
IB=mB(vB−uB)I_{B}=m_{B}\left(v_{B}-u_{B}\right)

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